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anonymous
 3 years ago
What is a quartic function with only the two real zeroes given?
x = –1 and x = –3
anonymous
 3 years ago
What is a quartic function with only the two real zeroes given? x = –1 and x = –3

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can go with (x + 1)^2 times (x + 3)^2 The reason this works is that it will give you a x^4 term and you still have only the 2 roots. Roots are of the form: (x  root) for you factors.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, basically you have both roots twice, so that's why each factor has a 2 on it for an exponent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You could also have put a 3 on one of them and left the other root as a single power (no exponent)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is this making sense to you?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0np, I'll give you a little more detail: First off, a quartic polynomial function has as its highest exponent, an x^4 term. That means that if you have an equation of the form: y = (x  root1)(x  root2)(x  root3)(x  root4) you will have quartic function. But we can have only 2 different roots, so a good choice is to have root1 twice and root 2 twice: y = [(x  root1)^2][(x  root2)^2]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's the same as: y = (x  root1)(x  root1)(x  root2)(x  root2) which also has root1 twice and root2 twice. It's just writing out the factors without the exponents. This equation will give you 0 for "y" if you replace x with either root, which is what you want to solve this problem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And to finish this off, you are given x = 1 and x = 3 as the only 2 roots you can have. And since the factor to give "0" using the root is: (x  root) one factor is [x  (1)] which is (x + 1) the other is [x  (3)] which is (x + 3) So, your polynomial is: y = (x + 1)(x + 1)(x + 3)(x + 3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You have other choices, but the above polynomial works. You could also have: y = (x + 1)(x + 3)(x + 3)(x + 3) or y = (x + 1)(x + 1)(x + 1)(x + 3) Both of those are quartics having only 2 different roots. That should make it all clearer now. Now are you good on this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here's a graph of the first solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorta so which one would it be? y = x4 – 4x3 – 4x2 – 4x – 3 y = –x4 + 4x3 + 4x2 + 4x + 3 y = x4 + 4x3 + 3x2 + 4x – 4 y = x4 + 4x3 + 4x2 + 4x +3 I understand how you did that i just don't know which one it will be.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I didn't know that you had choices. That changes the whole way you approach this problem. You can solve it 2 ways that I know of. The first way rst and the best and fastest is to perfrom synthetic division on each choice until you hit upon the choice that has only those 2 roots.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Number #1 is out because 1 is not a root. So, I eliminated that one so far.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Number 2 is out because it also has +1 as a root.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you sure you wrote those choices down correctly?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I actually copied and paste so yeah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And the roots are supposed to be 1 and 3 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah that is why i couldn't figure it out. That is what confused me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It looks like like the last one works. I tried it again.
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